JEE Main 2019MathematicsFunctionsEasyMCQ

JEE Main 2019 — Functions Question with Solution

JEE Main 2019 (09 Jan Shift 2)

Question

Let f:0,1→R  be such that fxy=fx.fy, for all x,y∈0,1,  and  f(0)≠0. If y=y(x) satisfies the differential equation, dydx=fx with y0=1 then y14+y34 is equal to:

Choose an option

Show full solutionCorrect option: C
Correct answer
C3

Step-by-step explanation

Given relation is fxy=fx. fy ∀ x, y∈R  .....(i)

On putting x= y=0, we get f0=f20⇒f0=0, 1 but

f0≠0

⇒f0=1,

Now if we put y=0 in (i), we get

fx=1

Hence dydx=1⇒y=x+c

⇒y=x+1 since y0=1

⇒y14+y34

=14+1+34+1=3.

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About this question

This is a previous-year question from JEE Main 2019, covering the Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.